# O

xamples

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ode

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### The Question

Recall that a stack is a last in first out data structure– data is always added to and removed from the top.

You need to implement a Stack data structure that has the following functions

• push(value) //to add elements (to the top of the stack)
• pop() // removes the top element and returns its value
• min() //returns the minmum element

All of the above functions should run in constant ( O(1) ) time.

No arrays or array methods should be used.

### The Approach

• need to implement a regular stack with 1 extra requirement– returning min() at any given time

• you can't just keep track of the minimum value each time you push data onto the stack (compare the data being pushed with the previous minimum and update the minimum if it is smaller)
• WHY?

Idea: have a separate structure that keeps track of the minimum for each node added. when you remove a node, remove the minimum that corresponds

Stack

Min

6

6

Stack

Min

6

6

7

6

Stack

Min

6

6

7

6

3

3

Stack

Min

6

6

7

6

3

3

2

2

Stack

Min

6

6

7

6

3

3

Stack

Min

6

6

7

6

Stack

Min

6

6

### A Solution

``````function Stack() {
this.top = null;
}

Stack.prototype.minimum = function() {
return this.minStack.peek();
}

Stack.prototype.push = function(val) {
var newNode = new Node(val);
if (!this.top) {
this.top = newNode;
this.minStack = new Stack();
this.minStack.push(val);
return;
}
newNode.next = this.top;
this.top = newNode;
if (this.minimum() > val)
this.minStack.push(val);
else
this.minStack.push(this.minimum());
};

Stack.prototype.pop = function() {
if (!this.top) return null;
var popped = this.top.val;
this.top = this.top.next;
if (this.minStack)
this.minStack.pop();
return popped;
};

Stack.prototype.peek = function() {
if (!this.top) return null;
return this.top.val;
};

function Node(val,next) {
this.val = val;
this.next = next;
}``````

### A  Better Solution

``````function Stack() {
this.top = null;
}

Stack.prototype.push = function(val) {
var newNode = new Node(val);
if (!this.top) {
this.top = newNode;
return;
}
newNode.next = this.top;
this.top = newNode;
};

Stack.prototype.pop = function() {
if (!this.top) return null;
var popped = this.top.val;
this.top = this.top.next;
return popped;
};

Stack.prototype.peek = function() {
if (!this.top) return null;
return this.top.val;
};

function Node(val,next) {
this.val = val;
this.next = next;
}

function MinStack() {
Stack.call(this);
this.minStack = new Stack();
}

MinStack.prototype.push = function(val) {
var newNode = new Node(val);
if (!this.top) {
this.top = newNode;
this.minStack.push(val);
return;
}
newNode.next = this.top;
this.top = newNode;
if (this.minStack.peek() > val)
this.minStack.push(val);
}

MinStack.prototype.pop = function() {
if (!this.top) return null;
var popped = this.top.val;
if (this.minStack.peek() === popped)
this.minStack.pop();
this.top = this.top.next;
return popped;
}

MinStack.prototype.min = function() {
return this.minStack.peek();
}``````

By Seema Ullal

# Minimal Stack

Technical interview practice problem

• 1,868